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vladimir1956 [14]
3 years ago
9

Suppose that a solution is tested for NO2 ion by adding 3M H2SO4 and heating, repeating this process until no further reaction o

ccurs. This solution
is then analyzed for NO3 ion by adding FeSO4 solution and more H2SO4, heating the test tube and its contents. The NO3 ion reaction is positive. In this case, is it accurate to state that NO3 ion was present in the original solution? Briefly explain. Write appropriate equations to support your answer.
Chemistry
1 answer:
Arada [10]3 years ago
8 0

Answer:

The correct equation will be "[Fe(H2O)5]²⁺ + NO → Fe(H2O)5NO]²+ (brown ring) ".

Explanation:

For NO₂, the addition of H₂SO₄ forms HNO₂ comprising NO gas as well as HNO₂, the equation will be:

⇒  NO₂⁻ + 2H⁺ →  HNO2

⇒  HNO₂ + H₂O + H⁺  →  HNO₃ + NO

For NO₃⁻,

NO₃⁻, Fe₂+ oxidized to Fe₃ + then releases NO gas in the existence of H₂SO₄

⇒  3Fe₂⁺ + 4H⁺ + NO₃⁻  →  3Fe₃⁺ + NO + 2H₂O

Brown ring forming establishes NO3- presence in the initial test sample,

⇒  [Fe(H2O)5]²⁺ + NO  →  Fe(H2O)5NO]²+ (brown ring)

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A chemist mixes 1.00 g CuCl2 with an excess of (NH4)2HPO4 in dilute aqueous solution . He measures the evolution of 670 J of hea
kaheart [24]

Answer:

\Delta H will be 90054 J

Explanation:

Number of moles = (mass)/(molar mass)

Molar mass of CuCl_{2} = 134.45 g/mol

So, 1.00 g of CuCl_{2} = \frac{1.00}{134.45}mol of CuCl_{2} = 0.00744 mol of CuCl_{2}

0.00744 mol of CuCl_{2} produces 670 J of heat

So, 1 mol of CuCl_{2} produces \frac{670}{0.00744}J of heat or 90054 J of heat

4 0
3 years ago
How many formula units are there in 450 g of Na2S04?
Mashutka [201]

Formula units in 450 g of Na_{2} So_{4} is 1.93 × 10²⁴ formula units.

<u>Explanation:</u>

First we have to find the number of moles in the given mass by dividing the mass by its molar mass as,

$\frac{450 g}{142.04 g/mol} =  3.2 moles

Now, we have to multiply the number of moles of Na₂SO₄ by the Avogadro's number, 6.022 × 10²³ formula units/mol, so we will get the number of formula units present in the given mass of the compound.

3.2 mol × 6.022 × 10²³ = 1.93 × 10²⁴ formula units.

So, 1.93 × 10²⁴ formula units is present in 450g of Na₂SO₄.

4 0
3 years ago
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8 0
2 years ago
What is [h3o ] in a solution of 0.075 m hno2 and 0.030 m nano2?
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Answer is: concentration of hydrogenium ions is 9,54·10⁻⁵ M.
c(HNO₂) = 0,075 M.
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This is buffer solution, so use <span>Henderson–Hasselbalch equation:
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pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M).
pH = 4,35 - 0,33.
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8 0
3 years ago
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